Business Calculus This document is attributed to Shana Calaway, Dale Hoffman and David Lippman Chapter OPEN ASSEMBLY EDITION OPEN ASSEMBLY editions of Open Tetbooks have been disaggregated into chapters in order to facilitate their full and seamless integration with courseware. Eperienced on the Open Assembly platform, this edition permits instructors and students to introduce supplemental materials of any media type directly at the chapter-level--in addition to the module and topic level. OPEN ASSEMBLY instructors and students are offered the opportunity to engage an immersive and unified user eperience of courseware and content, within a single interface. www.openassembly.com This OPEN ASSEMBLY edition is adapted with no changes to the original content. Open Tetbook Store URL: http://www.opentetbookstore.com
Business Calculus Edition Shana Calaway Dale Hoffman David Lippman This book is also available to read free online at http://www.opentetbookstore.com/buscalc/
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Introduction Business Calculus Introduction A Preview of Calculus Calculus was first developed more than three hundred years ago by Sir Isaac Newton and Gottfried Leibniz to help them describe and understand the rules governing the motion of planets and moons. Since then, thousands of other men and women have refined the basic ideas of calculus, developed new techniques to make the calculations easier, and found ways to apply calculus to problems besides planetary motion. Perhaps most importantly, they have used calculus to help understand a wide variety of physical, biological, economic and social phenomena and to describe and solve problems in those areas. Part of the beauty of calculus is that it is based on a few very simple ideas. Part of the power of calculus is that these simple ideas can help us understand, describe, and solve problems in a variety of fields. About this book Chapter Review contains review material that you should recall before we begin calculus. Chapter The Derivative builds on the precalculus idea of the slope of a line to let us find and use rates of change in many situations. Chapter The Integral builds on the precalculus idea of the area of a rectangle to let us find accumulated change in more complicated and interesting settings. Chapter 4 Functions of Two Variables etends the calculus ideas of chapter to functions of more than one variable. Supplements An online course framework is available on MyOpenMath.com for this book. The course framework features: Links to individual sections of the e-tet. Overview videos. Algorithmic, auto-grading online homework for each section of the tet. Most problems have video help tied to the question. A collection of printable resources created by Shana Calaway for the Open Course Library project.
Introduction Business Calculus 4 How is Business Calculus Different? Students who plan to go into science, engineering, or mathematics take a year-long sequence of classes that cover many of the same topics as we do in our one-quarter or one-semester course. Here are some of the differences: No trigonometry We will not be using trigonometry at all in this course. The scientists and engineers need trigonometry frequently, and so a great deal of the engineering calculus course is devoted to trigonometric functions and the situations they can model. The applications are different The scientists and engineers learn how to apply calculus to physics problems, such as work. They do a lot of geometric applications, like finding minimum distances, volumes of revolution, or arclengths. In this class, we will do only a few of these (distance/velocity problems, areas between curves). On the other hand, we will learn to apply calculus in some economic and business settings, like maimizing profit or minimizing average cost, finding elasticity of demand, or finding the present value of a continuous income stream. These are applications that are seldom seen in a course for engineers. Fewer theorems, no proofs The focus of this course is applications rather than theory. In this course, we will use the results of some theorems, but we won t prove any of them. When you finish this course, you should be able to solve many kinds of problems using calculus, but you won t be prepared to go on to higher mathematics. Less algebra In this class, you will not need clever algebra. If you need to solve an equation, it will either be relatively simple, or you can use technology to solve it. In most cases, you won t need eact answers; calculator numbers will be good enough.
Introduction Business Calculus 5 Simplification and Calculator Numbers When you were in tenth grade, your math teacher may have impressed you with the need to simplify your answers. I m here to tell you she was wrong. The form your answer should be in depends entirely on what you will do with it net. In addition, the process of simplifying, often messy algebra, can ruin perfectly correct answers. From the teacher s point of view, simplifying obscures how a student arrived at his answer, and makes problems harder to grade. Moral: don t spend a lot of etra time simplifying your answer. Leave it as close to how you arrived at it as possible. When should you simplify?. Simplify when it actually makes your life easier. For eample, in Chapter it s easier to find a second derivative if you simplify the first derivative.. Simplify your answer when you need to match it to an answer in the book. You may need to do some algebra to be sure your answer and the book answer are the same. When you use your calculator A calculator is required for this course, and it can be a wonderful tool. However, you should be careful not to rely too strongly on your calculator. Follow these rules of thumb:. Estimate your answers. If you epect an answer of about 4, and your calculator says 500, you ve made an error somewhere.. Don t round until the very end. Every time you make a calculation with a rounded number, your answer gets a little bit worse.. When you answer an applied problem, find a calculator number. It doesn t mean much to suggest that the company should produce 00(.4) items; it s much more.5 meaningful to report that they should produce about 06 items. 4. When you present your final answer, round it to something that makes sense. If you ve found an amount of US money, round it to the nearest cent. If you ve computed the number of people, round to the nearest person. If there s no obvious contet, show your teacher at least two digits after the decimal place. 5. Occasionally in this course, you will need to find the eact answer. That means not a calculator approimation. (You can still use your calculator to check your answer.)
Introduction Business Calculus 6 Table of Contents Chapter : Review... 7 Section : Functions... 7 Section : Operations on Functions... 9 Section : Linear Functions... Section 4: Eponents... 4 Section 5: Quadratics... 46 Section 6: Polynomials and Rational Functions... 5 Section 7: Eponential Functions... 60 Section 8: Logarithmic Functions... 67 Chapter : The Derivative... 7 Section : Instantaneous Rate of Change and Tangent Lines... 74 Section : Limits and Continuity... 79 Section : The Derivative... 85 Section 4: Rates in Real Life... 9 Section 5: Derivatives of Formulas... 99 Section 6: Second Derivative and Concavity... 6 Section 7: Optimization... Section 8: Curve Sketching... 5 Section 9: Applied Optimization... 4 Section 0: Other Applications... 5 Section : Implicit Differentiation and Related Rates... 56 Chapter : The Integral... 6 Section : The Definite Integral... 6 Section : The Fundamental Theorem and Antidifferentiation... 8 Section : Antiderivatives of Formulas... 88 Section 4: Substitution... 95 Section 5: Additional Integration Techniques... 0 Section 6: Area, Volume, and Average Value... 05 Section 7: Applications to Business... Section 8: Differential Equations... 0 Chapter 4: Functions of Two Variables... Section : Functions of Two Variables... Section : Calculus of Functions of Two Variables... 5 Section : Optimization... 6 Table of Integrals... 7
Chapter Review Business Calculus 7 Chapter : Review Section : Functions What is a Function? The natural world is full of relationships between quantities that change. When we see these relationships, it is natural for us to ask If I know one quantity, can I then determine the other? This establishes the idea of an input quantity, or independent variable, and a corresponding output quantity, or dependent variable. From this we get the notion of a functional relationship in which the output can be determined from the input. For some quantities, like height and age, there are certainly relationships between these quantities. Given a specific person and any age, it is easy enough to determine their height, but if we tried to reverse that relationship and determine height from a given age, that would be problematic, since most people maintain the same height for many years. Function Function: A rule for a relationship between an input, or independent, quantity and an output, or dependent, quantity in which each input value uniquely determines one output value. We say the output is a function of the input. Eample In the height and age eample above, is height a function of age? Is age a function of height? In the height and age eample above, it would be correct to say that height is a function of age, since each age uniquely determines a height. For eample, on my 8 th birthday, I had eactly one height of 69 inches. However, age is not a function of height, since one height input might correspond with more than one output age. For eample, for an input height of 70 inches, there is more than one output of age since I was 70 inches at the age of 0 and. Function Notation To simplify writing out epressions and equations involving functions, a simplified notation is often used. We also use descriptive variables to help us remember the meaning of the quantities in the problem. Rather than write height is a function of age, we could use the descriptive variable h to represent height and we could use the descriptive variable a to represent age. height is a function of age if we name the function f we write h is f of a or more simply h f(a) we could instead name the function h and write h(a) which is read h of a This chapter was remied from Precalculus: An Investigation of Functions, (c) 0 David Lippman and Melonie Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter Review Business Calculus 8 Remember we can use any variable to name the function; the notation h(a) shows us that h depends on a. The value a must be put into the function h to get a result. Be careful - the parentheses indicate that age is input into the function (Note: do not confuse these parentheses with multiplication!). Function Notation The notation output f(input) defines a function named f. This would be read output is f of input Eample A function N f(y) gives the number of police officers, N, in a town in year y. What does f(005) 00 tell us? When we read f(005) 00, we see the input quantity is 005, which is a value for the input quantity of the function, the year (y). The output value is 00, the number of police officers (N), a value for the output quantity. Remember Nf(y). So this tells us that in the year 005 there were 00 police officers in the town. Tables as Functions Functions can be represented in many ways: Words (as we did in the last few eamples), tables of values, graphs, or formulas. Represented as a table, we are presented with a list of input and output values. This table represents the age of children in years and their corresponding heights. While some tables show all the information we know about a function, this particular table represents just some of the data available for height and ages of children. (input) a, age in years 5 5 6 7 8 9 0 (output) h, height inches 40 4 44 47 50 5 54 Eample Which of these tables define a function (if any)? Input Output 5 8 6 Input Output - 5 0 4 5 Input Output 0 5 5 4 The first and second tables define functions. In both, each input corresponds to eactly one output. The third table does not define a function since the input value of 5 corresponds with two different output values.
Chapter Review Business Calculus 9 Solving and Evaluating Functions: When we work with functions, there are two typical things we do: evaluate and solve. Evaluating a function is what we do when we know an input, and use the function to determine the corresponding output. Evaluating will always produce one result, since each input of a function corresponds to eactly one output. Solving equations involving a function is what we do when we know an output, and use the function to determine the inputs that would produce that output. Solving a function could produce more than one solution, since different inputs can produce the same output. Eample 4 Using the table shown, where Qg(n) a) Evaluate g() n Q 8 6 7 4 6 5 8 Evaluating g() (read: g of ) means that we need to determine the output value, Q, of the function g given the input value of n. Looking at the table, we see the output corresponding to n is Q7, allowing us to conclude g() 7. b) Solve g(n) 6 Solving g(n) 6 means we need to determine what input values, n, produce an output value of 6. Looking at the table we see there are two solutions: n and n 4. When we input into the function g, our output is Q 6 When we input 4 into the function g, our output is also Q 6 Graphs as Functions Oftentimes a graph of a relationship can be used to define a function. By convention, graphs are typically created with the input quantity along the horizontal ais and the output quantity along the vertical. Eample 5 Which of these graphs defines a function yf()?
Chapter Review Business Calculus 0 Looking at the three graphs above, the first two define a function yf(), since for each input value along the horizontal ais there is eactly one output value corresponding, determined by the y-value of the graph. The rd graph does not define a function yf() since some input values, such as, correspond with more than one output value. Vertical Line Test The vertical line test is a handy way to think about whether a graph defines the vertical output as a function of the horizontal input. Imagine drawing vertical lines through the graph. If any vertical line would cross the graph more than once, then the graph does not define only one vertical output for each horizontal input. Evaluating a function using a graph requires taking the given input and using the graph to look up the corresponding output. Solving a function equation using a graph requires taking the given output and looking on the graph to determine the corresponding input. Eample 6 Given the graph below, a) Evaluate f() b) Solve f() 4 a) To evaluate f(), we find the input of on the horizontal ais. Moving up to the graph gives the point (, ), giving an output of y. So f() b) To solve f() 4, we find the value 4 on the vertical ais because if f() 4 then 4 is the output. Moving horizontally across the graph gives two points with the output of 4: (-,4) and (,4). These give the two solutions to f() 4: - or This means f(-)4 and f()4, or when the input is - or, the output is 4. Notice that while the graph in the previous eample is a function, getting two input values for the output value of 4 shows us that this function is not one-to-one. Formulas as Functions When possible, it is very convenient to define relationships using formulas. If it is possible to epress the output as a formula involving the input quantity, then we can define a function.
Chapter Review Business Calculus Eample 7 Epress the relationship n + 6p as a function p f(n) if possible. To epress the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p [something involving n]. n + 6p 6p - n subtract n from both sides divide both sides by 6 and simplify n n p n 6 6 6 Having rewritten the formula as p, we can now epress p as a function: p f( n) n Not every relationship can be epressed as a function with a formula. As with tables and graphs, it is common to evaluate and solve functions involving formulas. Evaluating will require replacing the input variable in the formula with the value provided and calculating. Solving will require replacing the output variable in the formula with the value provided, and solving for the input(s) that would produce that output. Eample 8 Given the function a) Evaluate k() b) Solve k(t) kt () t + a) To evaluate k(), we plug in the input value into the formula wherever we see the input variable t, then simplify k () + k () 8 + So k() 0 b) To solve k(t), we set the formula for k(t) equal to, and solve for the input value that will produce that output k(t) substitute the original formula kt () t + t + subtract from each side t take the cube root of each side t
Chapter Review Business Calculus When solving an equation using formulas, you can check your answer by using your solution in the original equation to see if your calculated answer is correct. We want to know if kt () is true when t. k + ( ) ( ) + which was the desired result. Basic Toolkit Functions There are some basic functions that it is helpful to know the name and shape of. We call these the basic "toolkit of functions." For these definitions we will use as the input variable and f() as the output variable. Toolkit Functions Linear Constant: f( ) Identity: f( ) c, where c is a constant (number) Absolute Value: f ( ) Power Quadratic: f ( ) Cubic: f ( ) Reciprocal: f( ) Reciprocal squared: f( ) Square root: f( ) Cube root: f( )
Chapter Review Business Calculus Graphs of the Toolkit Functions Constant Function: f( ) Identity: f( ) Absolute Value: f ( ) Quadratic: f Cubic: ( ) f ( ) Square root: f( ) Cube root: f( ) Reciprocal: f( ) Reciprocal squared: f( ) One of our main goals in mathematics is to model the real world with mathematical functions. In doing so, it is important to keep in mind the limitations of those models we create.
Chapter Review Business Calculus 4 This table shows a relationship between circumference and height of a tree as it grows. Circumference, c.7.5 5.5 8..7 Height, h 4.5 45. 54.6 9. While there is a strong relationship between the two, it would certainly be ridiculous to talk about a tree with a circumference of - feet, or a height of 000 feet. When we identify limitations on the inputs and outputs of a function, we are determining the domain and range of the function. Domain and Range Domain: The set of possible input values to a function Range: The set of possible output values of a function Eample 9 Using the tree table above, determine a reasonable domain and range. We could combine the data provided with our own eperiences and reason to approimate the domain and range of the function h f(c). For the domain, possible values for the input circumference c, it doesn t make sense to have negative values, so c > 0. We could make an educated guess at a maimum reasonable value, or look up that the maimum circumference measured is about 9 feet. With this information we would say a reasonable domain is 0 < c 9 feet. Similarly for the range, it doesn t make sense to have negative heights, and the maimum height of a tree could be looked up to be 79 feet, so a reasonable range is 0 < h 79 feet. A more compact alternative to inequality notation is interval notation, in which intervals of values are referred to by the starting and ending values. Curved parentheses are used for strictly less than, and square brackets are used for less than or equal to. Since infinity is not a number, we can t include it in the interval, so we always use curved parentheses with and -. The table below will help you see how inequalities correspond to interval notation: Inequality Interval notation 5 < h 0 (5, 0] 5 h < 0 [5, 0) 5 < h < 0 (5, 0) h < 0 (,0) h 0 [0, ) all real numbers (, )
Chapter Review Business Calculus 5 Eample 0 Describe the intervals of values shown on the line graph below using set builder and interval notations. To describe the values,, that lie in the intervals shown above we would say, is a real number greater than or equal to and less than or equal to, or a real number greater than 5. As an inequality it is: or > 5 In interval notation: [,] (5, ) Eample Find the domain of each function: a) f ( ) + 4 b) g( ) 6 a) Since we cannot take the square root of a negative number, we need the inside of the square root to be non-negative. + 4 0 when 4. The domain of f() is [ 4, ). b) We cannot divide by zero, so we need the denominator to be non-zero. 6 0 when, so we must eclude from the domain. The domain of g() is (,) (, ).. Eercises. The amount of garbage, G, produced by a city with population p is given by G f( p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces tons of garbage each week. Epress this information in terms of the function f. b. Eplain the meaning of the statement f ( 5).. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D ga ( ). a. A garden with area 5000 ft requires 50 cubic yards of dirt. Epress this information in terms of the function g. b. Eplain the meaning of the statement g ( 00).
Chapter Review Business Calculus 6. Select all of the following graphs which represent y as a function of. a b c d e f 4. Select all of the following graphs which represent y as a function of. a b c d e f 5. Select all of the following tables which represent y as a function of. a. 5 0 5 b. 5 0 5 c. 5 0 0 y 8 4 y 8 8 y 8 4 6. Select all of the following tables which represent y as a function of. a. 6 b. 6 6 c. 6 y 0 0 y 0 4 y 0 4
Chapter Review Business Calculus 7 7. Given the function g ( ) graphed here, a. Evaluate g () b. Solve g( ) 8. Given the function f( ) graphed here. a. Evaluate f ( 4) b. Solve f( ) 4 9. Based on the table below, a. Evaluate f () b. Solve f( ) 0 4 5 6 7 8 9 f( ) 74 8 5 56 6 45 4 47 0. Based on the table below, a. Evaluate f (8) b. Solve f( ) 7 0 4 5 6 7 8 9 f( ) 6 8 7 8 86 7 70 9 75 4 For each of the following functions, evaluate: f ( ), f ( ), f (0), f (), and f (). f ( ) 4. f ( ) 8. f ( ) 8 7 + 4. f ( ) 6 7 + 4 5. f ( ) + + 6. ( ) 7. f ( ) 8. ( ) + 9. Let f ( t) t+ 5 a. Evaluate f (0) b. Solve f ( t ) 0 f 4 f 0. Let g( p) 6 p a. Evaluate g (0) b. Solve g( p ) 0. Using the graph shown, a. Evaluate f() c b. Solve f ( ) p a b t r L c f() c. What are the coordinates of points L and K? K p
Chapter Review Business Calculus 8. Match each graph with its equation. i. ii. iii. iv. a. y b. y c. y d. y e. y f. y g. y h. y v. vi. vii. viii. Write the domain and range of each graph as an inequality.. 4. Find the domain of each function 5. f ( ) 6. f ( ) 5 + 9 6 7. f ( ) 8. f ( ) 6 8 9. f ( ) + 4 + 0. f ( ) 5 + 4
Chapter Review Business Calculus 9 Section : Operations on Functions Composition of Functions Suppose we wanted to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The temperature depends on the day, and the cost depends on the temperature. Using descriptive variables, we can notate these two functions. The first function, C(T), gives the cost C of heating a house when the average daily temperature is T degrees Celsius, and the second, T(d), gives the average daily temperature of a particular city on day d of the year. If we wanted to determine the cost of heating the house on the 5 th day of the year, we could do this by linking our two functions together, an idea called composition of functions. Using the function T(d), we could evaluate T(5) to determine the average daily temperature on the 5 th day of the year. We could then use that temperature as the input to the C(T) function to find the cost to heat the house on the 5 th day of the year: C(T(5)). Composition of Functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. We write f(g()), and read this as f of g of or f composed with g at. An alternate notation for composition uses the composition operator: ( f g)( ) is read f of g of or f composed with g at, just like f(g()). Eample Suppose c(s) gives the number of calories burned doing s sit-ups, and s(t) gives the number of sit-ups a person can do in t minutes. Interpret c(s()). When we are asked to interpret, we are being asked to eplain the meaning of the epression in words. The inside epression in the composition is s(). Since the input to the s function is time, the is representing minutes, and s() is the number of sit-ups that can be done in minutes. Taking this output and using it as the input to the c(s) function will gives us the calories that can be burned by the number of sit-ups that can be done in minutes. Composition of Functions using Tables and Graphs When working with functions given as tables and graphs, we can look up values for the functions using a provided table or graph. We start evaluation from the provided input, and first evaluate the inside function. We can then use the output of the inside function as the input to the outside function. To remember this, always work from the inside out. This chapter was remied from Precalculus: An Investigation of Functions, (c) 0 David Lippman and Melonie Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter Review Business Calculus 0 Eample Using the graphs below, evaluate f( g ()). g() f() To evaluate f( g ()), we again start with the inside evaluation. We evaluate g () using the graph of the g() function, finding the input of on the horizontal ais and finding the output value of the graph at that input. Here, g (). Using this value as the input to the f function, f( g()) f(). We can then evaluate this by looking to the graph of the f() function, finding the input of on the horizontal ais, and reading the output value of the graph at this input. Here, f () 6, so f( g ()) 6. Composition using Formulas When evaluating a composition of functions where we have either created or been given formulas, the concept of working from the inside out remains the same. First we evaluate the inside function using the input value provided, then use the resulting output as the input to the outside function. Eample Given f ( t) t t and h ( ) +, evaluate f( h ()). Since the inside evaluation is h() we start by evaluating the h() function at : h ( ) () + 5 Then f( h()) f(5), so we evaluate the f(t) function at an input of 5: f ( h()) f (5) 5 5 0 We are not limited, however, to using a numerical value as the input to the function. We can put anything into the function: a value, a different variable, or even an algebraic epression, provided we use the input epression everywhere we see the input variable.
Chapter Review Business Calculus Eample 4 Let f ( ) and g( ), find f(g()) and g(f()). To find f(g()), we start by evaluating the inside, writing out the formula for g() g( ) We then use the epression as input for the function f. f ( g( )) f We then evaluate the function f() using the formula for g() as the input. Since f ( ) then f This gives us the formula for the composition: f ( g( )) Likewise, to find g(f()), we evaluate the inside, writing out the formula for f() g ( f ( )) g( ) Now we evaluate the function g() using as the input. g( f ( )) Eample 5 A city manager determines that the ta revenue, R, in millions of dollars collected on a population of p thousand people is given by the formula R ( p) 0. 0p + p, and that the city s population, in thousands, is predicted to follow the formula p ( t) 60 + t + 0.t, where t is measured in years after 00. Find a formula for the ta revenue as a function of the year. Since we want ta revenue as a function of the year, we want year to be our initial input, and revenue to be our final output. To find revenue, we will first have to predict the city population, and then use that result as the input to the ta function. So we need to find R(p(t)). Evaluating this, ( 60 + t + 0.t ) 0.0( 60 + t + 0.t ) + 60 + t 0. R ( p( t)) R + t This composition gives us a single formula which can be used to predict the ta revenue during a given year, without needing to find the intermediary population value.
Chapter Review Business Calculus For eample, to predict the ta revenue in 07, when t 7 (because t is measured in years after 00) R( p(7)) 0.0 60 ( + (7) + 0.(7) ) + 60 + (7) + 0.(7). 079 million dollars Later in this course, it will be desirable to decompose a function to write it as a composition of two simpler functions. Eample 6 Write f ( ) 5 + as the composition of two functions. We are looking for two functions, g and h, so f ( ) g( h( )). To do this, we look for a function inside a function in the formula for f(). As one possibility, we might notice that 5 is the inside of the square root. We could then decompose the function as: h( ) 5 g ( ) + We can check our answer by recomposing the functions: g( h( )) g 5 + 5 ( ) Note that this is not the only solution to the problem. Another non-trivial decomposition would be h ( ) and g( ) + 5 Transformations of Functions Transformations allow us to construct new equations from our basic toolkit functions. The most basic transformations are shifting the graph vertically or horizontally. Vertical Shift Given a function f(), if we define a new function g() as g ( ) f( ) + k, where k is a constant then g() is a vertical shift of the function f(), where all the output values have been increased by k. If k is positive, then the graph will shift up If k is negative, then the graph will shift down Horizontal Shift Given a function f(), if we define a new function g() as g ( ) f( + k), where k is a constant then g() is a horizontal shift of the function f() If k is positive, then the graph will shift left
Chapter Review Business Calculus If k is negative, then the graph will shift right Eample 7 Given f( ), sketch a graph of h ( ) f ( + ) +. The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f( + ) is a change on the inside of the function, giving a horizontal shift left by, then the subtraction by in f( + ) is a change to the outside of the function, giving a vertical shift down by. Transforming the graph gives Eample 8 Write a formula for the graph shown, a transformation of the toolkit square root function. The graph of the toolkit function starts at the origin, so this graph has been shifted to the right, and up. In function notation, we could write that as h ( ) f( ) +. Using the formula for the square root function we can write h ( ) + Note that this transformation has changed the domain and range of the function. This new graph has domain [, ) and range [, ). Another transformation that can be applied to a function is a reflection over the horizontal or vertical ais.
Chapter Review Business Calculus 4 Reflections Given a function f(), if we define a new function g() as g ( ) f( ), then g() is a vertical reflection of the function f(), sometimes called a reflection about the -ais If we define a new function g() as g ( ) f( ), then g() is a horizontal reflection of the function f(), sometimes called a reflection about the y-ais Eample 9 A common model for learning has an equation similar to t kt () +, where k is the percentage of mastery that can be achieved after t practice sessions. This is a transformation of the function f() t t shown here. Sketch a graph of k(t). This equation combines three transformations into one equation. A horizontal reflection: f( t) t combined with A vertical reflection: f( t) t combined with A vertical shift up : t f( t) + + We can sketch a graph by applying these transformations one at a time to the original function: The original graph Horizontally reflected Then vertically reflected Then, after shifting up, we get the final graph: t kt () f( t) + +. Note: As a model for learning, this function would be limited to a domain of t 0, with corresponding range [0,).
Chapter Review Business Calculus 5 With shifts, we saw the effect of adding or subtracting to the inputs or outputs of a function. We now eplore the effects of multiplying the outputs. Vertical Stretch/Compression Given a function f(), if we define a new function g() as g ( ) kf ( ), where k is a constant then g() is a vertical stretch or compression of the function f(). If k >, then the graph will be stretched If 0< k <, then the graph will be compressed If k < 0, then there will be combination of a vertical stretch or compression with a vertical reflection Eample 0 The graph to the right is a transformation of the toolkit function f ( ). Relate this new function g() to f(), then find a formula for g(). When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g ( ). With the basic cubic function at the same input, f () 8. Based on that, it appears that the outputs of g are ¼ the outputs of the function f, since g ( ) f (). From this 4 we can fairly safely conclude that: g ( ) f ( ) 4 We can write a formula for g by using the definition of the function f g ( ) f ( ) 4 4 Combining Transformations When combining vertical transformations, it is very important to consider the order of the transformations. For eample, vertically shifting by and then vertically stretching by does not create the same graph as vertically stretching by and then vertically shifting by. The order follows nicely from order of operations. Combining Vertical Transformations When combining vertical transformations written in the form af ( ) + k,
Chapter Review Business Calculus 6 first vertically stretch by a, then vertically shift by k. Eample Write an equation for the transformed graph of the quadratic function shown. Since this is a quadratic function, first consider what the basic quadratic tool kit function looks like and how this has changed. Observing the graph, we notice several transformations: The original tool kit function has been flipped over the ais, some kind of stretch or compression has occurred, and we can see a shift to the right units and a shift up unit. In total there are four operations: Vertical reflection, requiring a negative sign outside the function Vertical Stretch Horizontal Shift Right units, which tells us to put - on the inside of the function Vertical Shift up unit, telling us to add on the outside of the function By observation, the basic tool kit function has a verte at (0, 0) and symmetrical points at (, ) and (-, ). These points are one unit up and one unit over from the verte. The new points on the transformed graph are one unit away horizontally but units away vertically. They have been stretched vertically by two. Not everyone can see this by simply looking at the graph. If you can, great, but if not, we can solve for it. First, we will write the equation for this graph, with an unknown vertical stretch. f ( ) The original function Vertically reflected f ( ) Vertically stretched af ( ) a af ( ) a( ) Shifted right af ( ) + a( ) + Shifted up We now know our graph is going to have an equation of the form g ( ) a( ) +. To find the vertical stretch, we can identify any point on the graph (other than the highest point), such as the point (,-), which tells us g ( ). Using our general formula, and substituting for, and - for g() a( ) + a + a a This tells us that to produce the graph we need a vertical stretch by two.
Chapter Review Business Calculus 7 The function that produces this graph is therefore g ( ) ( ) +. Eample On what interval(s) is the function g ( ) + increasing and decreasing? ( ) This is a transformation of the toolkit reciprocal squared function, f ( ) : f ( ) A vertical flip and vertical stretch by f ( ) ( ) A shift right by f ( ) + + A shift up by ( ) The basic reciprocal squared function is increasing on (,0) and decreasing on ( 0, ). Because of the vertical flip, the g() function will be decreasing on the left and increasing on the right. The horizontal shift right by will also shift these intervals to the right one. From this, we can determine g() will be increasing on (, ) and decreasing on (,). We also could graph the transformation to help us determine these intervals.. Eercises ( ) ( ) Given each pair of functions, calculate f g ( 0) and g f ( 0).. f ( ) 4+ 8, g( ) 7. f ( ) 5+ 7, g( ) 4. f ( ) + 4, g( ) 4. f ( ), g ( ) 4 + + Use the table of values to evaluate each epression 5. f( g (8)) 6. f ( g ( 5) ) 7. g( f (5)) 8. g( f ( ) ) 9. f( f (4)) 0. f ( f ( ) ). gg ( ()) g g 6 ( ). ( ) f( ) g ( ) 0 7 9 6 5 5 6 8 4 4 5 0 8 6 7 7 8 9 4 9 0
Chapter Review Business Calculus 8 Use the graphs to evaluate the epressions below.. f( g ()) 4. f ( g ( ) ) 5. g( f ()) 6. g( f ( 0) ) 7. f( f (5)) 8. f ( f ( 4) ) 9. gg ( ()) g g 0 ( ) 0. ( ) ( ) ( ) For each pair of functions, find f g( ) and ( ) 6 7 6. f ( ), g ( ) +. ( ) g f. Simplify your answers. f 4 g + 4. f ( ) +, g( ) + 4. f ( ) +, ( ) 5. f ( ), g( ) 5+ 6. f ( ), g( ) 7. If ( ) 4 g + f + 6, g ( ) 6 and h ( ), find f( gh ( ( ))) 8. If f ( ) +, g( ) and h( ) +, find f( gh ( ( ))) + 9. The function D( p ) gives the number of items that will be demanded when the price is p. The production cost, C ( ) is the cost of producing items. To determine the cost of production when the price is $6, you would do which of the following: a. Evaluate DC ( (6)) b. Evaluate CD ( (6)) c. Solve DC ( ( )) 6 d. Solve CDp ( ( )) 6 0. The function Ad ( ) gives the pain level on a scale of 0-0 eperienced by a patient with d milligrams of a pain reduction drug in their system. The milligrams of drug in the patient s system after t minutes is modeled by mt (). To determine when the patient will be at a pain level of 4, you would need to: 4 m A 4 ( ) ( ) 4 ( ) ( ) 4 a. Evaluate Am ( ) b. Evaluate ( ) c. Solve Amt ( ) d. Solve mad ( )
Chapter Review Business Calculus 9 ( ) Find functions f( ) and g ( ) so the given function can be epressed as h( ) f g( ). h( ) ( + ). h( ) ( 5) 5. h( ) 4. h ( ) ( + ) 5. h( ) + 6. h( ) 4 + Sketch a graph of each function as a transformation of a toolkit function. 7. f ( t) ( t+ ) 8. h( ) + 4 9. k( ) ( ) 40. mt ( ) + t+ 4. f ( ) 4( + ) 5 4. g ( ) 4. h( ) 4 + 44. k( ) Write an equation for each function graphed below. 4 ( ) 5 +. 45. 46. 47. 48. 49. 50.
Chapter Review Business Calculus 0 For each function graphed, estimate the intervals on which the function is increasing and decreasing. 5. 5.
Chapter Review Business Calculus Section : Linear Functions As you hop into a taicab in Allentown, the meter will immediately read $.0; this is the drop charge made when the taimeter is activated. After that initial fee, the taimeter will add $.40 for each mile the tai drives. In this scenario, the total tai fare depends upon the number of miles ridden in the tai, and we can ask whether it is possible to model this type of scenario with a function. Using descriptive variables, we choose m for miles and C for Cost in dollars as a function of miles: C(m). We know for certain that C ( 0). 0, since the $.0 drop charge is assessed regardless of how many miles are driven. Since $.40 is added for each mile driven, we could write that if m miles are driven, C( m).0 +. 40m because we start with a $.0 drop fee and then for each mile increase we add $.40. It is good to verify that the units make sense in this equation. The $.0 drop charge is measured in dollars; the $.40 charge is measured in dollars per mile. So dollars C( m).0dollars +. 40 ( m miles) mile When dollars per mile are multiplied by a number of miles, the result is a number of dollars, matching the units on the.0, and matching the desired units for the C function. Notice this equation C( m).0 +. 40m consisted of two quantities. The first is the fied $.0 charge which does not change based on the value of the input. The second is the $.40 dollars per mile value, which is a rate of change. In the equation this rate of change is multiplied by the input value. Looking at this same problem in table format we can also see the cost changes by $.40 for every mile increase. m 0 C(m).0 5.70 8.0 0.50 It is important here to note that in this equation, the rate of change is constant; over any interval, the rate of change is the same. Graphing this equation, C( m).0 +. 40m we see the shape is a line, which is how these functions get their name: linear functions When the number of miles is zero the cost is $.0, giving the point (0,.0) on the graph. This is the vertical or C(m) intercept. The graph is increasing in a straight line from left to right because for each mile the cost goes up by $.40; this rate remains consistent. This chapter was remied from Precalculus: An Investigation of Functions, (c) 0 David Lippman and Melonie Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter Review Business Calculus Linear Function A linear function is a function whose graph produces a line. Linear functions can always be written in the form f ( ) b + m or f ( ) m + b ; they re equivalent where b is the initial or starting value of the function (when input, 0), and m is the constant rate of change of the function This form of a line is called slope-intercept form of a line. Many people like to write linear functions in the form corresponds to the way we tend to speak: The output starts at b and increases at a rate of m. f ( ) b + m because it For this reason alone we will use the f ( ) b + m form for many of the eamples, but remember they are equivalent and can be written correctly both ways. Slope and Increasing/Decreasing m is the constant rate of change of the function (also called slope). The slope determines if the function is an increasing function or a decreasing function. f ( ) b + m is an increasing function if m > 0 f ( ) b + m is a decreasing function if m < 0 If m 0, the rate of change zero, and the function f( ) b+ 0 b is just a horizontal line passing through the point (0, b), neither increasing nor decreasing. Eample Marcus currently owns 00 songs in his itunes collection. Every month, he adds 5 new songs. Write a formula for the number of songs, N, in his itunes collection as a function of the number of months, m. How many songs will he own in a year? The initial value for this function is 00, since he currently owns 00 songs, so N ( 0) 00. The number of songs increases by 5 songs per month, so the rate of change is 5 songs per month. With this information, we can write the formula: N( m) 00 + 5m. N(m) is an increasing linear function. With this formula we can predict how many songs he will have in year ( months): N ( ) 00 + 5() 00 + 80 80. Marcus will have 80 songs in months.
Chapter Review Business Calculus Calculating Rate of Change Given two values for the input, and, and two corresponding values for the output, y and y, or a set of points, (, y) and (, y ), if we wish to find a linear function that contains both points we can calculate the rate of change, m: change in output y y y m change in input Rate of change of a linear function is also called the slope of the line. Note in function notation, y f ) and y f ), so we could equivalently write m ( ) ( ) f f ( ( Eample The population of a city increased from,400 to 7,800 between 00 and 006. Find the rate of change of the population during this time span. The rate of change will relate the change in population to the change in time. The population increased by 7800 400 4400 people over the 4 year time interval. To find the rate of change, the number of people per year the population changed by: 4400 people people 00 00 people per year 4years year Notice that we knew the population was increasing, so we would epect our value for m to be positive. This is a quick way to check to see if your value is reasonable. Eample The pressure, P, in pounds per square inch (PSI) on a diver depends upon their depth below the water surface, d, in feet, following the equation P( d) 4.696 + 0. 44d. Interpret the components of this function. output pressure PSI The rate of change, or slope, 0.44 would have units. This tells us input depth ft the pressure on the diver increases by 0.44 PSI for each foot their depth increases. The initial value, 4.696, will have the same units as the output, so this tells us that at a depth of 0 feet, the pressure on the diver will be 4.696 PSI.
Chapter Review Business Calculus 4 We can now find the rate of change given two input-output pairs, and could write an equation for a linear function if we had the rate of change and initial value. If we have two input-output pairs and they do not include the initial value of the function, then we will have to solve for it. Eample 4 Write an equation for the linear function graphed to the right. Looking at the graph, we might notice that it passes through the points (0, 7) and (4, 4). From the first value, we know the initial value of the function is b 7, so in this case we will only need to calculate the rate of change: m 4 7 4 0 4 This allows us to write the equation: f ( ) 7 4 Eample 5 If f () is a linear function, f ( ), and f ( 8), find an equation for the function. In eample, we computed the rate of change to be m. In this case, we do not know the 5 initial value f (0), so we will have to solve for it. Using the rate of change, we know the equation will have the form f ( ) b +. Since we know the value of the function when 5, we can evaluate the function at. f ( ) b + () 5 Since we know that f ( ), we can substitute on the left side b + () 5 This leaves us with an equation we can solve for the initial value 9 9 b 5 5 Combining this with the value for the rate of change, we can now write a formula for this function: 9 f ( ) + 5 5
Chapter Review Business Calculus 5 As an alternative to the approach used above to find the initial value, b, we can use the pointslope form of a line instead. Point-Slope Equation of a Line An equation for the line passing through the point (, y ) with slope m can be written as y y m( ) This is called the point-slope form of a line. It is a little easier to write if you know a point and the slope, but requires a bit of work to rewrite into slope-intercept form, and requires memorizing another formula. Eample 6 Working as an insurance salesperson, Ilya earns a base salary and a commission on each new policy, so Ilya s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold new policies, and earned $760 for the week. The week before, he sold 5 new policies, and earned $90. Find an equation for I(n), and interpret the meaning of the components of the equation. The given information gives us two input-output pairs: (,760) and (5,90). We start by finding the rate of change. 90 760 60 m 80 5 Keeping track of units can help us interpret this quantity. Income increased by $60 when the number of policies increased by, so the rate of change is $80 per policy; Ilya earns a commission of $80 for each policy sold during the week. We can now write the equation using the point-slope form of the line, using the slope we just found and the point (,760): I 760 80( n ) If we wanted this in function form (slope intercept form), we could rewrite the equation into that form: I 760 80( n ) I 760 80n 40 I( n) 50 + 80n This form allows us to see the starting value for the function: 50. This is Ilya s income when n 0, which means no new policies are sold. We can interpret this as Ilya s base salary for the week, which does not depend upon the number of policies sold. Our final interpretation is: Ilya s base salary is $50 per week and he earns an additional $80 commission for each policy sold each week.